I did a scan of Math.SE and couldn't find this, so I will post:
I am working on Bass's Analysis exercise 2.3 in page 15 (on book) http://bass.math.uconn.edu/3rd.pdf
The problem states, given:
$$ A_1 \subset A_2 \subset A_3 ... $$ where each finite $A_i$ is a $\sigma$ -algebra, prove that $\bigcup_{\forall i} A_i $ is a $\sigma$-algbra or find a counter example.
Now I was able to show that: $\bigcup_{\forall i} A_i $ is a set algebra by considering that any finite collection of set $u_1 ... u_m \in \bigcup_{\forall i} A_i$ must belong to $A_{u_1} ,... A_{u_m}$ respectively and therefore there is a finite indexed $A_k$ that contains $A_{u_1} ,... A_{u_m}$ and therefore all the countable unions and intersections of $u_1 ... u_m \in \bigcup_{\forall i}$ must be contained in $A_k$ (and similarly the complements of the sets).
Now this argument dies if you consider a countably infinite sequence of sets $u_1, u_2 , u_3...$ since you can no longer talk about the "largest, finite-index m such that $A_m$ contains $u_1, u_2 ... $ "
Which is causing me start to doubt that $\bigcup_{\forall i} A_i $ is even a $\sigma$ algebra in the first place. Does anyone know how to find a counterexample? Or how to modify the proof to circumvent this obstacle.
